A paper weight is dropped from the roof of a 35 storey building each storey being 3m high. It passes the ceiling of the 20th storey at x m/s [g=10 m/s2]. Find x.

To solve the problem of finding the speed of the paperweight as it passes the ceiling of the 20th storey, we can use the principles of physics, specifically the equations of motion under constant acceleration due to gravity. Let's break this down step by step.

Understanding the Scenario

We know that the building has 35 storeys, each 3 meters high. Therefore, the height of the building can be calculated as:

• Total height = Number of storeys × Height per storey
• Total height = 35 × 3 m = 105 m

We are interested in the speed of the paperweight when it reaches the ceiling of the 20th storey. The height of the 20th storey can be calculated as:

• Height of the 20th storey = 20 × 3 m = 60 m

Applying the Equations of Motion

When the paperweight is dropped, it starts from rest, which means its initial velocity (u) is 0 m/s. The acceleration (a) due to gravity is given as 10 m/s². We can use the following equation of motion to find the final velocity (v) when it reaches the height of 60 m:

v² = u² + 2as

In this equation:

• v = final velocity (what we want to find)
• u = initial velocity = 0 m/s
• a = acceleration = 10 m/s²
• s = distance fallen = 60 m

Calculating the Final Velocity

Now, substituting the known values into the equation:

v² = 0² + 2 × 10 × 60

v² = 0 + 1200

v² = 1200

To find v, we take the square root of both sides:

v = √1200

v = 34.64 m/s

Final Result

Thus, the speed of the paperweight as it passes the ceiling of the 20th storey is approximately 34.64 m/s.

This example illustrates how we can apply the equations of motion to determine the speed of an object in free fall, taking into account the height it has fallen from and the acceleration due to gravity. If you have any further questions or need clarification on any part of this process, feel free to ask!